Question:
Evaluate the following integral:
$\int \frac{x^{2}+9}{x^{4}+81} d x$
Solution:
re-writing the given equation as
$\int \frac{1+\frac{9}{x^{2}}}{x^{2}+\frac{81}{x^{2}}} d x$
$\int \frac{1+\frac{9}{x^{2}}}{\left(x-\frac{9}{x}\right)^{2}+18} d x$
Let $x-\frac{9}{x}=t$
$\left(1+\frac{9}{x^{2}}\right) d x=d t$
$\int \frac{d t}{t^{2}+18}$
Using identity $\int \frac{1}{x^{2}+1} d x=\arctan (x)$
$\frac{1}{3 \sqrt{2}} \arctan \left(\frac{t}{3 \sqrt{2}}\right)+c$
Substituting $\mathrm{t}$ as $\mathrm{x}-\frac{1}{\mathrm{x}}$
$\frac{1}{3 \sqrt{2}} \arctan \left(\frac{x-\frac{1}{x}}{3 \sqrt{2}}\right)+c$